Notes on Units and Physical Quantities

The Nature of Physics

Physics is an experimental science in which physicists seek patterns in several natural phenomena.
The patterns are called physical theories.
A very well established or widely used theory is called a physical law or principle.
This course introduces how to describe and analyze physical phenomena using quantities, units, and measurements.

A physical quantity is any number that is used to describe a physical phenomenon.
Examples of physical quantities:
- Force: F = 30\;\text{N}
- Time: 60\;\text{s}
- Length: 1.0\;\text{m}
- Mass: 50\;\text{kg}
The term “standard magnitude” is used to illustrate a representative numerical value for a quantity.

The International System (SI base units) is also known as the metric system.
Seven base quantities and their units:
- Length: name = \text{meter}, symbol = \text{m}
- Mass: name = \text{kilogram}, symbol = \text{kg}
- Time: name = \text{second}, symbol = \text{s}
- Electric current: name = \text{ampere}, symbol = \text{A}
- Thermodynamic temperature: name = \text{kelvin}, symbol = \text{K}
- Amount of substance: name = \text{mole}, symbol = \text{mol}
- Luminous intensity: name = \text{candela}, symbol = \text{cd}
These base units form the foundation from which all other (derived) units are constructed.

A physical equation must be dimensionally consistent: terms that are added or equated must have the same units.
Always compare similar quantities (apples to apples).
Example of an incorrect dimensional match: \text{Mass} \neq \text{Length}
Dimensional analysis helps verify the consistency of equations and the dimensions of quantities such as velocity, acceleration, etc.
For example, if velocity is expressed as a product of acceleration and time, then the dimensions must satisfy:
- Velocity = acceleration × time
- [v] = [a][t] = \left(\frac{\text{m}}{\text{s}^2}\right)(\text{s}) = \frac{\text{m}}{\text{s}}
- Hence, [v] = [L][T]^{-1} and [a] = [L][T]^{-2}.

Uses of dimensional analysis:
- Derive an equation.
- Check if an equation is dimensionally correct.
- Determine the units or the dimension of a physical quantity.
- Check/simplify the dimension of the LHS and RHS of an equation.

Problem: Determine if the following are dimensionally correct:
1) x = vt
2) v = m + 2ax
3) x = vt + \frac{1}{2}at^2
Given symbols:
- x has dimension [L]
- m has dimension [M]
- v has dimension \left[\frac{L}{T}\right]
- t has dimension [T]
- a has dimension \left[\frac{L}{T^2}\right]

1) x = vt
- LHS: [x] = [L]
- RHS: [v][t] = \left[\frac{L}{T}\right][T] = [L]
- Therefore, dimensionally correct.
2) v = m + 2ax
- LHS: [v] = [\frac{L}{T}]
- RHS: [m] + 2[a][x] = [M] + \left[\frac{L}{T^2}\right][L] = [M] + [L^2 T^{-2}]
- Mismatch: cannot add [M] to [L^2 T^{-2}]; dimensionally incorrect.
3) x = vt + \frac{1}{2}at^2
- RHS terms: [v][t] = [L],\; \frac{1}{2}[a][t]^2 = \frac{1}{2}\left[\frac{L}{T^2}\right][T^2] = [L]
- Sum of two length terms; LHS: [x] = [L]; dimensionally correct if the sum of two L quantities is meaningful.

Example: 18 years old ⇒ how many seconds?
Given: 18 years ≈ 568,036,800 seconds; exactly 5.680368 \times 10^{8} \text{ s}
Conversion path (typical):
- 1 year ≈ 365.25 days
- 1 day = 24 hours; 1 hour = 3600 seconds
Expression: 18\ \text{years} \times \frac{365.25\ \text{days}}{\text{year}} \times \frac{24\ \text{h}}{\text{day}} \times \frac{3600\ \text{s}}{\text{h}} = 5.680368 \times 10^{8} \text{ s}

Final answers should be expressed with the number of significant figures corresponding to the given quantities.
Example: 5 s.f. (as a guideline in the course).
Example conversions:
- 1.0 \times 10^{4} \text{ m} has 2 significant figures.
- 1.04 \times 10^{4} \text{ m} has 3 significant figures.
- 10\,351\ \text{m} has 5 significant figures.

Scenarios illustrating uncertainty vs. measurement practice:
- A cardboard thickness measured with a ruler is recorded as 3\text{ mm}, not 3.00\text{ mm} (uncertainty not shown as extra decimals).
- A vendor does not quote a precise weight like 2.41735\text{ kg}; instead, a reasonable display of uncertainty/precision is used.
Distinctions:
- Accuracy: closeness to the true value.
- Precision: reproducibility or consistency among repeated measurements.
Visual intuition often shown as a 2x2 grid: Low/High accuracy vs Low/High precision (examples omitted here; concept explained).

Common representations of measurement uncertainty:
- 56.47 \pm 0.02\ \text{ mm}
- 56.47(2)\ \text{ mm} (the (2) indicates the uncertainty in the last digits; here ±0.02 mm)
- 47\,\Omega \pm 10\%

When uncertainties are not specified, use the number of meaningful digits (significant figures).
Example rules:
- 2.91\text{ mm} \Rightarrow 3\text{ sig figs}
- The last digit is in the hundredths place; uncertainty about 0.01\text{ mm}.
- 137\text{ km} \Rightarrow 3\text{ sig figs}; uncertainty about \sim 1\text{ km}.

All non-zero digits are significant.
Zeros before or after a decimal point are significant only if preceded by a non-zero digit.
If there is no decimal point, zeros to the right of the rightmost non-zero digit are not significant.
Examples:
- 1200 \quad\text{(2 sig figs)}
- 1200.0 \quad\text{(5 sig figs)}
- 13.20 \quad\text{(4 sig figs)}
- 112000 \quad\text{(3 sig figs)}
- 112000. \quad\text{(6 sig figs)}
- 0.0034560 \quad\text{(5 sig figs)}

Multiplication/Division: the result should have the same number of significant figures as the factor with the fewest significant figures.
- Example: \frac{(0.745)(2.2)}{3.885} = 0.424… \Rightarrow 0.42 (2 SF)
Addition/Subtraction: the result should have the same number of decimal places as the quantity with the fewest digits to the right of the decimal point.
- Example: 27.153 + 138.2 - 11.74 = 153.613 \Rightarrow 153.6 (1 decimal place)

University Physics 13th Ed, H. Young and R. Freedman, Pearson Education 2014
PowerPoint Lectures for University Physics 13th Ed, Wayne Anderson, Pearson Education 2012
Physics 71 Lectures by J Vance, A Lacaba, P J Blancas, G Pedemonte
Annotations by: Mark Ivan Ugalino
Edited by: Rene L. Principe Jr.